It is known that: $$\lim_{k\rightarrow\infty} H_k-\log(k)=\gamma$$Where $\gamma$ is Euler's constant. What if we change the base from $e$ to any other number $b$? How do we prove that this converges/diverges? I realized that for $b=3$ it seems to converge but then starts to grow slowly.
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1$\begingroup$ 1) Surely you mean $\log(k)$ instead of $\log(x)$. 2) Note that $\log_b(k) = \frac{\log k}{\log b}$. So if $b\neq e,$ the difference $H_k - \log_b(k)$ must diverge - for large $k$, this is $\gamma + o(1) + (1-1/\log(b))\cdot \log k$ which blows up with $k$. $\endgroup$– stochasticboy321Dec 25, 2022 at 23:19
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Note that $\log(k)= \int\limits_1^k\frac{1}{x}\mathrm{d}x$, so $H_{k-1}$ is a upper sum of that Integral and $\sum\limits_{n=2}^{k}\frac{1}{n}=H_k-1$ is a lower sum. Therefore we can bound $H_k-\log{(k)}$ via $H_k-H_{k-1}\le H_k-\log{(k)}\le H_k-(H_k-1) \Leftrightarrow \frac{1}{k}\le H_k-\log(k)\le 1$.
Thus your sequences always converges.
Edit: I kinda misread the question, for different bases, the limit will diverge because, as pointed out in the comments $H_k-\log_b(k)=-\log_b(k)+\log(k)+\gamma + \frac{1}{2k}+o(\frac{1}{k^2})$, which diverges
